Problem: $4lm + 2ln + 6l - 8 = 4m + 3$ Solve for $l$.
Explanation: Combine constant terms on the right. $4lm + 2ln + 6l - {8} = 4m + {3}$ $4lm + 2ln + 6l = 4m + {11}$ Notice that all the terms on the left-hand side of the equation have $l$ in them. $4{l}m + 2{l}n + 6{l} = 4m + 11$ Factor out the $l$ ${l} \cdot \left( 4m + 2n + 6 \right) = 4m + 11$ Isolate the $l$ $l \cdot \left( {4m + 2n + 6} \right) = 4m + 11$ $l = \dfrac{ 4m + 11 }{ {4m + 2n + 6} }$